Question:

A stone is thrown up vertically and the height $'x'$ ft reached by it in time $'t'$ secs is given by $x = 80t - 16t^2$ . The stone reaches the maximum height in time ?

Updated On: Jun 21, 2022
  • $2 \,secs$
  • $2.5\, secs$
  • $3\, secs$
  • $3.5\, secs$
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The Correct Option is B

Solution and Explanation

We have , $ x = 80 t - 16t^2$
$ \therefore \:\:\: \frac{dx}{dt} = 80 - 32 t$
Now , $\frac{dx}{dt} = 0 \Rightarrow 80 - 32 t = 0$
$ \Rightarrow \:\: t = \frac{5}{2}$
Now, $ \frac{d^2 x}{dt^2} = - 332 \: \Rightarrow \frac{d^2 x}{dt^2}|_{t = \frac{5}{2}} = - 32 < 0$
$ \Rightarrow \:\: t = \frac{5}{2}$ is maximum point
Hence, the stone reaches the maximum height in time, $ t = \frac{5}{2} = 2.5 $ secs
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives